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Tomtit [17]
3 years ago
9

David has some marbles.He gave 2/5 of his marble to his neighbour and 1/6 of the remainder to his friend Ali.He had 25 marbles l

eft.How many marbles did he have at first?
Mathematics
1 answer:
babunello [35]3 years ago
7 0

Answer: he had 50 marbles at first.

Step-by-step explanation:

Let x represent the number of marbles that he had at first.

He gave 2/5 of his marble to his neighbour. It means that the number of marbles that he gave to his neighbour is 2/5 × x = 2x/5

The number of marbles that he has remaining is

x - 2x/5 = 3x/5

He gave 1/6 of the remainder to his friend Ali. It means that the number of marbles that he gave to his friend is

1/6 × 3x/5 = x/10

The number of marbles that he has left is

3x/5 - x/10 = 5x/10

He had 25 marbles left. It means that

5x/10 = 25

5x = 25 × 10 = 250

x = 250/5 = 50

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3 0
3 years ago
HELP DUE TODAY!!!!!!
larisa [96]

Answer:

x° is 66°  

Step-by-step explanation:

From the given diagram, we have;

∠JIH = 105° Given

∠IDJ = 39° Given

Therefore, we have;

∠JID and ∠JIH are supplementary angles, by the sum of angles on a straight line

∴ ∠JID + ∠JIH = 180° by definition of supplementary angles

∠JID + 105° = 180° by substitution property

∠JID = 180° - 105° = 75° by angle subtraction postulate

∠JID = 75°

∠IDJ + ∠JID + ∠IJD = 180° by the sum of interior angles of a triangle

∠IJD = 180° - (∠IDJ + ∠JID)  = 180° - (39° + 75°) = 66° angle subtraction postulate

∠IJD = 66°

∠x° ≅ ∠IJD, by vertically opposite angles

∴ ∠x° = ∠IJD = 66° by the definition of congruency

∠x° = 66°  

7 0
3 years ago
Read 2 more answers
At the spring dance there were 23 more students than 6 times the number of teachers. There were 215 students that attended the d
Mademuasel [1]

Answer: There were 32 teachers.

Step-by-step explanation:

(215 - 23)/6 = amount of teachers

215 - 23 = 192

192/6 = 32

8 0
3 years ago
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A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is current
lapo4ka [179]

Using the z-distribution, we have that:

a)

The hypothesis test is:

  • H_0: \mu = 7.4
  • H_1: \mu < 7.4

The p-value is of 0.0668.

The critical value is z^{\ast} = -1.28

b) Since the <u>test statistic is less than the critical value</u> for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

Item a:

At the null hypothesis, it is <u>tested if the mean is of 7.4 minutes</u>, that is:

H_0: \mu = 7.4

At the alternative hypothesis, it is <u>tested if the mean is lower than 7.4 minutes</u>, that is:

H_1: \mu < 7.4

We have the <u>standard deviation for the population</u>, hence, the z-distribution is used.

The test statistic is given by:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • \sigma is the standard deviation of the sample.
  • n is the sample size.

For this problem, the values of the <em>parameters </em>are: \overline{x} = 7.1, \mu = 7.4, \sigma = 1.2, n = 36.

Hence, the value of the <em>test statistic</em> is:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{7.1 - 7.4}{\frac{1.2}{\sqrt{36}}}

z = -1.5

Using a z-distribution calculator, the p-value is of 0.0668.

Also using a calculator, considering a <u>left-tailed test</u>, as we are testing if the mean is less than a value, the critical value with a <u>significance level of 0.1</u> is of z^{\ast} = -1.28.

Item b:

Since the <u>test statistic is less than the critical value</u> for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

You can learn more about the use of the z-distribution to test an hypothesis at brainly.com/question/16313918

5 0
2 years ago
A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of sta
nalin [4]

Answer:

Option 2 - Approximately 24–36 pounds

Step-by-step explanation:

Given : A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of standard Eskimo dogs are normally distributed.

To find : What range of weights would 99.7% of the dogs have?

Solution :

The range of 99.7% will lie between the mean ± 3 standard deviations.

We have given,

Mean weight of Eskimo dogs is \mu=30

Standard deviation of Eskimo dogs is \sigma=2

The range of weights would 99.7% of the dogs have,

R=\mu\pm3\sigma

R=30\pm3(2)

R=30\pm6

R=30+6,30-6

R=36,24

Therefore, The range is approximately, 24 - 36 pounds.

So, Option 2 is correct.

5 0
4 years ago
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